Question 980190: 1.Julie Ann has some loom bands to give to her friends.If she gives 6 or 8 loom bands to each friend,she has no left over.If she gives 9 to each friend,she has 3 left ove.Find the smallest number of loom bands Julie Ann can have.
2.Find a three digit number such that the hundreds digit and the unit digit are the same prime numbers,the sum of all the digit is 12 and the number is divisible by 7.
3.What is the sum of the digits of 10 raise to the power of 2014-36?
4.The LCM of two integers a and b is 4,200 and their GCF is 100.If neither a nor b is multiple of the other,find a+b.
5.How many unique rectangles with integral sides can be formed whose area is 2014 square units?
thanks a lot.
Answer by onlinepsa(22)   (Show Source):
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1.Julie Ann has some loom bands to give to her friends.If she gives 6 or 8 loom bands to each friend,she has no left over.If she gives 9 to each friend,she has 3 left ove.Find the smallest number of loom bands Julie Ann can have.
Ans: Let, number of loom bands be 6k1 or 8k2.
6k1= 8k2
We need to find an expression that satisfies both the conditions:
Number pattern for 6k1 : 6, 12, 18, 24, 30, 42, 48, 54, 60, 66, 72....
Number pattern for 8k2 : 8, 16, 24 , 32, 40, 48, 56, 64, 72, ...
Observing the common elements, we can observe the series: 24, 48 , 72....
We can write the expression to be 24k. [We can also arrive here by taking LCM of 6 and 8 which is 24]
Also, 24k when divided by 9 gives a remainder 3. Thus, we can write:
24k = 9x + 3 [x is the quotient when 24k is divided by 9].
Rearranging the above equation:
9x = 24k - 3
=> 9x = 18k + (6k -3)
LHS is a multiple of 9. RHS should also be. Since 18k in RHS is already a multiple of 9, (6k -3) should also be.
By trial and error, k=2 makes 6k-3 a multiple of 9. x comes out to be 5.
Thus, the smallest possible number of loom bands = 9x +3 = 48.
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2.Find a three digit number such that the hundreds digit and the unit digit are the same prime numbers,the sum of all the digit is 12 and the number is divisible by 7.
Ans:- Let the three digit number be 'XYZ'.
X=Z = could be 2, 3, 5 or 7 according to the question. In fact, 7 is ruled out as the sum of all digits is 12.
i) Let X= Z= 2
X+Z=4; Y= 8; 484 is however not divisible by 7.
Let X= Z= 3, X+Z=6; Y= 6; 363 is not divisible by 7.
Let X= Z= 5, X+Z=10; Y= 2; 525 is divisible by 7.
Ans-> 525.
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3.What is the sum of the digits of 10 raise to the power of 2014-36?
Ans:- Let us observe the pattern first..
10^2 - 36 = 64 [Zero 9s, one 6, one 4]
10^3 - 36 = 964 [One 9s, one 6, one 4]
10^4 - 36 = 9964 [Two 9s, one 6, one 4]
..
..
10^2014 - 36 = 9999....64 [Two thousand twelve 9s, one 6, one 4]
Thus, sum of all the digits= 2012*9 + 6 + 4 = 18,108 + 10 = 18,118.
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4.The LCM of two integers a and b is 4,200 and their GCF is 100.If neither a nor b is multiple of the other,find a+b.
Ans:
Since, the GCF = 100, let the numbers be 100x and 100y such that x and y do not have any common factors (other than 1).
a + b = 100 (x + y)
LCM = 4200
=> 100 * x * y = 4200
=> xy = 42
(x,y) that satisfy the above equation= (1,42) , (2,21) ,(3,14) , (6,7)
(1,42) set is being rejected based upon the condition "neither a nor b is multiple of the other".
However, in the rest three case all seem to be satisfying the condition.
Possible sum= 100 (2+21) , 100 (3+14), 100 (6+7)
=2300, 1700, 1300
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5.How many unique rectangles with integral sides can be formed whose area is 2014 square units?
Ans: Let the sides be l and b.
l*b=2014= 2*1007= 2*19*53
Unique rectangles: (1,2014) , (2,1007) , (19, 106) , (37, 53)
By unique rectangles, we can understand that (19,106) and (106,19) are one and the same as they can be superimposed over each other.
4 unique rectangles.
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Thanks,
PRD
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